Answer:
The maximum height of the ball is 256m
Step-by-step explanation:
Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t
At maximum height, the velocity of the ball is zero.
velocity = dh/dt
velocity = -32t + 128
Since v = 0 at maximum height
0 = -32t+128
32t = 128
t = 128/32
t = 4seconds
The maximum height can be gotten by substituting t = 4 into the modelled equation.
h = -16t² + 128t
h = -16(4)²+128(4)
h = -16(16)+512
h = -256+512
h = 256m
If u are trying to solve for x is that x=2
4 in (4,-5) implies positive x-axis.
-5 in (4,-5) implies negative y-axis.
Since x is positive and y is negative in fourth quadrant, so the answer is
QUADRANT IV (or 4th quadrant.
Answer:
m = -3m+3n/-2b
Step-by-step explanation:
Slope is expressed as;
m = y2-y1/x2-x1
Given the coordinates (a+b,4m-n) and (a-b,m+2n)
m = (m+2n)-(4m-n)/(a-b)-(a+b)
m = m+2n-4m+n/a-b-a-b
m = m-4m+2n+n/a-a-b-b
m = -3m+3n/-2b
Hence the required slope is m = -3m+3n/-2b
